【LeetCode 49】字母异位词分组

比较复杂的方法。

    public static List<List<String>> groupAnagrams(String[] strs) {
    	List<List<String>> list = new ArrayList<List<String>>();
    	int n = strs.length;
    	if (n == 0)
    		return null;
    	int[] pointers = new int[n+1];
    	Arrays.fill(pointers, -2);
    	int i = 0;
    	int j = i+1;
    	int k = 1; // the position in pointers 
    	pointers[0] = -1;
    	while (i < n)
    	{
    		while (j < n) // 相邻两个字符串的等价情况
        	{
        		if (isAnagram(strs[i], strs[j]))
        		{
        			i++;
        			j++;
        		}
        		else
        			break;
        	}
        	while (j < n && !isAnagram(strs[i], strs[j])) // 寻找不相邻字符串的等价情况
        		j++;
        	if (j < n) // 使得其相邻
        	{
        		exchange(strs, i+1, j);
        		i++;
        		//j++;
        	}
        	if (j == n)
        	{
        		pointers[k++] = i;
        		i++;
        		j = i+1;
        	}
    	}
    	
    	int p = 0; // position in pointer array
    	int t = 0; 
    	while (p + 1 <= n && pointers[p] != -2)
    	{
    		List<String> ret = new ArrayList<String>();
    		// pointers[p]+1 到 pointers[p+1]是一个区间的位置
    		// p+1应该位于pointers数组之中
        	for (t = pointers[p]+1; t <= pointers[p+1]; t++)
        	{
        		ret.add(strs[t]);
        	}
        	if (!ret.isEmpty())
        		list.add(ret);
        	p++;
    	}
    	return list;
    }
    
    private static void exchange(String[] strs, int i, int j)
    {
    	String tmp = strs[i];
    	strs[i] = strs[j];
    	strs[j] = tmp;
    }
    
    private static boolean isAnagram(String s1, String s2) // O(n)
    {
    	boolean ret = false;
    	int n1 = s1.length();
    	int n2 = s2.length();
    	if (n1 != n2)
    		return ret;
    	Map<Character, Integer> charMap = new HashMap<Character, Integer>();
    	for (int i = 0; i < n1; i++)
    	{
    		char c = s1.charAt(i);
    		if (charMap.containsKey(c))
    			charMap.put(c, charMap.get(c)+1);
    		else
    			charMap.put(c, 1);
    	}
    	for (int i = 0; i < n2; i++)
    	{
    		char c = s2.charAt(i);
    		if (!charMap.containsKey(c))
    			return ret;
    		if (charMap.containsKey(c) && charMap.get(c) > 0)
    		{
    			charMap.put(c, charMap.get(c)-1);
    			if (charMap.get(c) == 0)
    				charMap.remove(c);
    		}
    	}
    	if (charMap.isEmpty())
    		ret = true;
    	return ret;
    }

标签：charMap,49,int,异位,pointers,++,ret,strs,LeetCode
来源： https://blog.csdn.net/juttajry/article/details/122255645